$S_4$ does not have a normal subgroup of order $8$

Question is to prove that :

$S_4$ does not have a normal subgroup of order $8$

I do not have any specific idea how to proceed but:

Assuming there exists a normal subgroup $H$ of order $8$ in $S_4$, As $H\unlhd G$, $HK\leq G$ for any subgroup $K\leq G$ and $|HK|=\frac{|H||K|}{|H\cap K|}$

what i am trying to do is try getting an element $x$ of order $2$ which is not in $H$ and set $K=\{1,x\}$ then $HK$ would be a group of order $16$ which is a contradiction as $S_4$ can not have a subgroup of order $16$.

as there are six $2-cycles$ and three products of disjoint $2$ cycles but $|H|=8$, there does exists an element of order $2$ which is not in $H$ and thus we are done.

I am sure this would be the nicest way or the stupidest way one can ever do :P

I would be thankful if someone can help me to see if anything is wrong in my approach.

I would be thankful if someone can give me a hint for an alternate approach.

Thank You :)

• This is in fact quite a nice approach in my opinion. – Bungo Oct 1 '17 at 2:32

You approach is OK, but here is a simpler one. Since $S_4/H \cong C_3$ is abelian, it follows that $[S_4,S_4]=A_4 \subseteq H$. Can you see this leads to a contradiction?

• +1. I do not understand how much more time will it take for me to write simple solutions than writing a one page solution..... :( I understand the contradiction, $|A_4|=12$ and $|H|=8$.. – user87543 Sep 5 '13 at 14:02
• Praphulla, do not worry! The beauty of mathematics is that different proofs can lead to the same theorem. This is how you learn mathenmatics! It happened many times that famous mathematicians went back to their/an original proof and simplified it substantially! This phenomenon warrants maybe a separate entry in this StackExchange ... – Nicky Hekster Sep 5 '13 at 14:22
• Thanks for your concern Sir, :) – user87543 Sep 6 '13 at 5:44
• @nickyhekster what does $[S_4,S_4]$ stand for? – Guacho Perez Apr 13 '17 at 15:44
• The commutator subgroup, so (isomorphic to) the alternating group $A_4$. In general $[G,G]=\langle x^{-1}y^{-1}xy : x,y \in G \rangle$. – Nicky Hekster Apr 13 '17 at 15:50

Yet another approach: Normal subgroups are comprised of entire conjugacy classes, one of which must be the identity's conjugacy class, but the class equation for $S_4$ is $$24=1+3+6+6+8,$$ and there is no way to obtain $8$ as a sum of terms from the right hand side including $1$.

• this is useful after knowing "class equation". but i am solving problems on cosets and lagrange theorem from the book "ABSTRACT ALGEBRA-DUMMIT FOOTE". So, I can not use this approach. But i liked it. :) – user87543 Sep 5 '13 at 14:12

Another approach : Two elements in $S_n$ are conjugate to each other iff they have the same cycle decomposition. Hence, all the transpositions must be in $H$ if one of them is (because $H$ is normal).

This leaves one non-identity element, which must have order 2 (since all the transpositions are their own inverses). That element must be a product of transpositions. Such a thing would be conjugate to something else, which is not in $H$.

If you assume that $H$ does not contain any transpositions, it still contains an element of order 2 ...

• at first sight i think i understood, but I am not sure now. Assuming $H$ has a transposition we concluded that all transpositions must be in $H$. This leaves one non identity element. I see that it has to be product of transpositions, and its conjugate has to be in $H$ which is not possible. So, H is not normal. right? – user87543 Sep 5 '13 at 14:16
• Yes, exactly. Same goes if you assume a product of two disjoint transpositions is in $H$ – Prahlad Vaidyanathan Sep 5 '13 at 16:02
• Could you please elaborate the case where $H$ does not contain transpositions. I see that it should still have an element of order 2, so an element of the type product of 2 transpositions. There are 3 of those. But then where is the contradiction? Thanks in advance – John11 Nov 2 '18 at 13:35

If you have the Sylow theorems available, then you can argue as follows. The dihedral group $D_4$ (symmetries of a square) has order $8$ and can be viewed as a subgroup of $S_4$ by regarding a symmetry of the square as a permutation of the four vertices. The permutation $p$ that interchanges two adjacent vertices of the square while fixing the other two vertices is in $S_4$ but not in $D_4$. Since it has order $2$, this $p$ must be in a $2$-Sylow subgroup. So you have at least two distinct $2$-Sylow subgroups of $S_4$, namely $D_4$ and one containing $p$. But all the $2$-Sylow subgroups are conjugate to each other, so, as soon as there's more than one of them, none of them can be normal.

• hayyo!! it is becoming more and more complicated :O – user87543 Sep 6 '13 at 5:43

Suppose $$N$$ is a normal subgroup of $$S_4$$ of order $$8$$. The $$2$$-cycles are: $$(12), (13), (14), (23), (24)$$ and $$(34)$$. So we have $$6$$ elements of order $$2$$.

Claim: There exists an element of order $$2$$ that is not in $$N$$.

Proof: If $$N$$ contains all the $$2$$-cycles of $$S$$, then $$1, (12), (13), (14), (23), (24)$$ and $$(34)$$ are all in $$N$$. Since $$|N|=8$$, then $$N$$ contains an element of order $$3$$ or $$4$$. So $$N$$ must also contain the inverse of that element (and the elements of order $$3$$ or $$4$$ are not self-inverse) so that $$N$$ will contain at least $$9$$ elements. This contradicts the order of $$N$$. Therefore there exists a $$2$$-cycle $$(ab)\notin N$$.

Now take $$H=\{1,(ab)\}$$. $$H$$ is a subgroup of $$G$$. Since $$N$$ is normal in $$G$$, $$HN$$ is a subgroup of $$G$$. Clearly, $$H\cap N=1$$ since $$(ab)\notin N$$. It follows that $$|HN|=\frac{|H||N|}{|H\cap N|}=|H||N|=16$$. But $$16$$ is not a divisor of $$24$$, contradiction Lagrange's Theorem. Therefore $$S_4$$ does not have a normal subgroup of order $$8$$.